What is the integral from 0 to 1 of x^2+y^2 ?

The integral from 0 to 1 of x^2+y^2 is y^2+1/3

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integral from 0 to 1 of x^2+y^2

\int_{0}^{1} x^{2} + y^{2} d x

y^{2} + \frac{1}{3}

Solution steps

\int_{0}^{1} x^{2} + y^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} y^{2} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} y^{2} d x = y^{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= y^{2} + \frac{1}{3}

\int_{1}^{3} \frac{7}{x} d x

\int_{0}^{1} 3 x d x

\int_{21}^{29} x x^{2} - 441 d x

\int_{0}^{0.1} \frac{1}{x ln ∣ x ∣} d x

\int_{0}^{1} (2 - x)^{2} d x

What is the integral from 0 to 1 of x^2+y^2 ?
The integral from 0 to 1 of x^2+y^2 is y^2+1/3